Chapter 4

To perform addition, subtraction, multiplication and division of complex numbers.

To understand the concept of the complex conjugate.

To represent complex numbers graphically on an Argand diagram.

To work with complex numbers in modulusargument form, and to understand the

geometric interpretation of multiplication and division of complex numbers in this form.To understand and apply De Moivres theorem.To factorise polynomial expressions over C and to solve polynomial equations over C.To sketch subsets of the complex plane, including lines, rays and circles.

FIN

III

To understand the imaginary number i and the set of complex numbers C.

AL

IIIIII

PAG

Objectives

ES

Complex numbers

In the sixteenth century, mathematicians including Girolamo Cardano began to consider

square roots of negative numbers. Although these numbers were regarded as impossible,they arose in calculations to find real solutions of cubic equations.For example, the cubic equation x3 15x 4 = 0 has three real solutions. Cardanos formulagives the solutionp3p3

x = 2 + 121 + 2 121

which you can show equals 4.

Today complex numbers are widely used in physics and engineering, such as in the study ofaerodynamics.

Let z = 4 5i. Find:

c Re(z) Im(z) = 4 (5) = 9

4A Starting to build the complex numbers 157

Using the TI-Nspire

the first line. Use to access i. To find the real part, use menu > Number >Complex Number Tools > Real Part, or justtype real(. For the imaginary part, use menu > Number> Complex Number Tools > Imaginary Part.

ES

Assign the complex number z, as shown in

Note: You do not need to be in complex mode. If you use i in the input, then it will

display in the same format.

PAG

Using the Casio ClassPad

In M, tap Real in the status bar at the bottom ofthe screen to change to Cplx mode. Enter 4 5i z and tap EXE .Note: The symbol i is found in the

the main screen).

160 Chapter 4: Complex numbers

I Argand diagramsAn Argand diagram is a geometric representation of the set of complex numbers. In a vectorsense, a complex number has two dimensions: the real part and the imaginary part. Thereforea plane is required to represent C.

32

(2 + i)

A complex number written as a + bi is said to

be in Cartesian form.

Example 5

(3 + i)

0 13 2 11

Re(z)

23

PAG

Each point on an Argand diagram represents a

complex number. The complex number a + biis situated at the point (a, b) on the equivalentCartesian axes, as shown by the examples inthis figure.

Represent the following complex numbers as points on an Argand diagram:

4A Starting to build the complex numbers 161

I Geometric representation of the basic operations on

complex numbersAddition of complex numbers is analogous to addition of vectors. The sum of two complexnumbers corresponds to the sum of their position vectors.Multiplication of a complex number by a scalar corresponds to the multiplication of itsposition vector by the scalar.Im(z)

ES

Im(z)

z1 + z2

az

z2

bz

Re(z)

Re(z)

a>10<b<1c<0

PAG

z1

cz

The difference z1 z2 is represented by the sum z1 + (z2 ).

Example 6

Let z1 = 2 + i and z2 = 1 + 3i.

Represent the complex numbers z1 , z2 , z1 + z2 and z1 z2 on an Argand diagram and show

162 Chapter 4: Complex numbers

I Multiplication of complex numbers

We carried out this calculation with an assumption that we are in a system where all the usualrules of algebra apply. However, it should be understood that the following is a definition ofmultiplication for C.Multiplication of complex numbers

Let z1 = a + bi and z2 = c + di. Then

PAG

z1 z2 = (ac bd) + (ad + bc)i

The multiplicative identity for C is 1 = 1 + 0i. The following familiar properties of the realnumbers extend to the complex numbers: z1 z2 = z2 z1

(z1 z2 )z3 = z1 (z2 z3 )

Example 7Simplify:a (2 + 3i)(1 5i)

b 3i(5 2i)

Solution

AL

a (2 + 3i)(1 5i) = 2 10i + 3i 15i 2

z1=z

z1 (z2 + z3 ) = z1 z2 + z1 z3

c i3

b 3i(5 2i) = 15i 6i 2

= 2 10i + 3i + 15

= 15i + 6

= 17 7i

= 6 + 15i

FIN

I Geometric significance of multiplication by i

When the complex number 2 + 3i is multiplied by 1,the result is 2 3i. This is achieved through a rotationof 180 about the origin.

c i3 = i i2= i

Im(z)

2 + 3i

3 + 2i

When the complex number 2 + 3i is multiplied by i,

we obtain

Re(z)

i(2 + 3i) = 2i + 3i 2= 2i 3= 3 + 2i

2 3i

The result is achieved through a rotation of 90 anticlockwise about the origin.

If 3 + 2i is multiplied by i, the result is 2 3i. This is again achieved through a rotationof 90 anticlockwise about the origin.Final pages Cambridge University Press Evans et al., 2015 ISBN 978-1-107-58743-4 Ph 03 8671 1400

Let z = 2 i. Simplify the following:

Prove that |z1 + z2 | |z1 | + |z2 | for all z1 , z2 C.

4C The modulusargument form of a complex number 169

4C The modulusargument form of a complex number

In the preceding sections, we have expressed complex numbers in Cartesian form. Anotherway of expressing complex numbers is using modulusargument (or polar) form.Each complex number may be described by an angle and a distance from the origin. In thissection, we will see that this is a very useful way to describe complex numbers.

The diagram shows the point P corresponding to the

complex number z = a + bi. We see that a = r cos andb = r sin , and so we can writez = a + bi= r cos + (r sin ) i

= r cos + i sin

ES

I Polar formIm(z)

P z = a + bi

Re(z)

PAG

This is called the polar form of the complex number. The polar form is abbreviated toz = r cis The distance r =

a2 + b2 is called the modulus of z and is denoted by |z|.

The angle , measured anticlockwise from the horizontal axis, is called the argument of zand is denoted by arg z.Polar form for complex numbers is also called modulusargument form.

FIN

AL

This Argand diagram uses a polar grid

= 15 .with rays at intervals of12

Im(z)

2cis 23

2i

2cis 56

2cis 3cis 3

2cis 62

cis 23

cis 3

Re(z)

2i

I Non-uniqueness of polar form

Each complex number has more than one representation in polar form.Since cos = cos( + 2n) and sin = sin( + 2n), for all n Z, we can writez = r cis = r cis( + 2n)

Find the complex conjugate of each of the following:

174 Chapter 4: Complex numbers

4D Basic operations on complex numbers in

modulusargument formI Addition and subtractionThere is no simple way to add or subtract complex numbers in the form r cis . Complexnumbers need to be expressed in the form a + bi before these operations can be carried out.

Now use the compound angle formulas from Chapter 3:

Here are two useful properties of the modulus and the principal argument with regard tomultiplication of complex numbers: |z1 z2 | = |z1 | |z2 |

Arg(z1 z2 ) = Arg(z1 ) + Arg(z2 ) + 2k, where k = 0, 1 or 1

AL

I Geometric interpretation of multiplication

We have seen that:

Im(z)

The modulus of the product of two complex

FIN

numbers is the product of their moduli.

The argument of the product of two complexnumbers is the sum of their arguments.

z1z2

Geometrically, the effect of multiplying a complex

number z1 by the complex number z2 = r2 cis 2 is toproduce an enlargement of Oz1 , where O is the origin,by a factor of r2 and an anticlockwise turn through anangle 2 about the origin.

r1r20

2 r11

z1Re(z)

If r2 = 1, then only the turning effect will take place.

Let z = cis . Multiplication by z2 is, in effect, the same as a multiplication by z followed byanother multiplication by z. The effect is a turn of followed by another turn of . The endresult is an anticlockwise turn of 2. This is also shown by finding z2 :z2 = z z = cis cis = cis( + )

4E Solving quadratic equations over the complex numbers 183

d From Example 19c, we have

Note: In parts a, b and c of this example, the two solutions are conjugates of each other.

ES

We explore this further in the next section.

Using the TI-Nspire

menu

> Algebra

PAG

To find complex solutions, use

> Complex > Solve as shown.

Using the Casio ClassPad

Ensure your calculator is in complex mode.

Enter and highlight the equation.

Select Interactive > Equation/Inequality > solve.

Ensure that the variable is z.

AL

We can see that any quadratic polynomial can be factorised into linear factors over thecomplex numbers. In the next section, we find that any higher degree polynomial can also befactorised into linear factors over the complex numbers.

Exercise 4E1

Factorise each of the following into linear factors over C:

FIN

Skillsheet

aceg

Example 18, 19

Example 20

z2 + 16z2 + 2z + 52z2 8z + 93z2 + 2z + 2

bdfh

z2 + 5z2 3z + 43z2 + 6z + 42z2 z + 3

Solve each of the following equations over C:

aceg

x2 + 25 = 0x2 4x + 5 = 0x2 = 2x 3z2 + (1 + 2i)z + (1 + i) = 0

bdfh

x2 + 8 = 03x2 + 7x + 5 = 05x2 + 1 = 3xz2 + z + (1 i) = 0

Hint: Show that 3 + 4i = (1 + 2i)2 .

184 Chapter 4: Complex numbers

4F Solving polynomial equations over the complex numbers

You have studied polynomials over the real numbers in Mathematical Methods Units 3 & 4.We now extend this study to polynomials over the complex numbers.For n N {0}, a polynomial of degree n is an expression of the formP(z) = an zn + an1 zn1 + + a1 z + a0

ES

where the coefficients ai are complex numbers and an , 0.

When we divide the polynomial P(z) by the polynomial D(z) we obtain two polynomials,Q(z) the quotient and R(z) the remainder, such thatP(z) = D(z)Q(z) + R(z)and either R(z) = 0 or R(z) has degree less than D(z).

PAG

If R(z) = 0, then D(z) is a factor of P(z).

The remainder theorem and the factor theorem are true for polynomials over C.Remainder theorem

Let C. When a polynomial P(z) is divided by z , the remainder is P().

Factor theorem

Let C. Then z is a factor of a polynomial P(z) if and only if P() = 0.

4F Solving polynomial equations over the complex numbers 185

I The conjugate root theorem

We have seen in the examples in this section and the previous section that, for polynomialequations with real coefficients, there are solutions which are conjugates.Conjugate root theorem

Let P(z) be a polynomial with real coefficients. If a + bi is a solution of the equation

P(z) = 0, with a and b real numbers, then the complex conjugate a bi is also a solution.Proof We will prove the theorem for quadratics, as it gives the idea of the general proof.Let P(z) = az2 + bz + c, where a, b, c R and a , 0. Assume that is a solution of theequation P(z) = 0. Then P() = 0. That is,a2 + b + c = 0

AL

Take the conjugate of both sides of this equation and use properties of conjugates:a2 + b + c = 0a2 + b + c = 0

a(2 ) + b + c = 0

since a, b and c are real numbers

a() + b + c = 0

FIN

Hence P() = 0. That is, is a solution of the equation P(z) = 0.

If a polynomial P(z) has real coefficients, then using this theorem we can say that thecomplex solutions of the equation P(z) = 0 occur in conjugate pairs.

I Factorisation of cubic polynomials

Over the complex numbers, every cubic polynomial has three linear factors.If the coefficients of the cubic are real, then at least one factor must be real (as complexfactors occur in pairs). The usual method of solution, already demonstrated in Example 21,is to find the real linear factor using the factor theorem and then complete the square on theresulting quadratic factor. The cubic polynomial can also be factorised if one complex rootis given, as shown in the next example.

I The fundamental theorem of algebra

the coefficients ai are complex numbers, has at least one linear factor in the complexnumber system.Given any polynomial P(z) of degree n 1, the theorem tells us that we can factorise P(z) asP(z) = (z 1 )Q(z)

for some 1 C and some polynomial Q(z) of degree n 1.

PAG

By applying the fundamental theorem of algebra repeatedly, it can be shown that:

A polynomial of degree n can be factorised into n linear factors in C:

A polynomial equation can be solved by first rearranging it into the form P(z) = 0, whereP(z) is a polynomial, and then factorising P(z) and extracting a solution from each factor.If P(z) = (z 1 )(z 2 ) . . . (z n ), then the solutions of P(z) = 0 are 1 , 2 , . . . , n .The solutions of the equation P(z) = 0 are also referred to as the zeroes or the roots of thepolynomial P(z).

4G Using De Moivres theorem to solve equations 191

4G Using De Moivres theorem to solve equations

Equations of the form zn = a, where a C, are often solved by using De Moivres theorem.Write both z and a in polar form, as z = r cis and a = r1 cis .Then zn = a becomes(r cis )n = r1 cis (using De Moivres theorem)

a Find the square roots of 1 + i by using:

b Hence find exact values of cos

4H Sketching subsets of the complex plane

Particular sets of points of the complex plane can be described by placing restrictions on z.For example:

z : Re(z) = 6 is the straight line parallel to the imaginary axis with each point on the linehaving real part 6.

z : Im(z) = 2 Re(z) is the straight line through the origin with gradient 2.The set of all points which satisfy a given condition is called the locus of the condition(plural loci). When sketching a locus, a solid line is used for a boundary which is included inthe locus, and a dashed line is used for a boundary which is not included.

Example 29

AL

On an Argand diagram, sketch the subset S of the complex plane, where

S = z : |z 1| = 2Solution

Im(z)

Method 1 (algebraic)

Let z = x + yi. Then

FIN

|z 1| = 2

|x + yi 1| = 2

|(x 1) + yi| = 2

1 + 0i

0 1 + 0i

3 + 0i

Re(z)

(x 1)2 + y2 = 2

(x 1)2 + y2 = 4

This demonstrates that S is represented by the circle with centre 1 + 0i and radius 2.

Method 2 (geometric)

If z1 and z2 are complex numbers, then |z1 z2 | is the distance between the points on thecomplex plane corresponding to z1 and z2 .

Hence z : |z 1| = 2 is the set of all points that are distance 2 from 1 + 0i. That is, the setS is represented by the circle with centre 1 + 0i and radius 2.

4H Sketching subsets of the complex plane 195

Example 30On an Argand diagram, sketch the subset S of the complex plane, where

S = z : |z 2| = |z (1 + i)|SolutionIm(z)

Method 1 (algebraic)

Let z = x + yi. Then

|x + yi 2| = |x + yi (1 + i)|

|x 2 + yi| = |x 1 + (y 1)i|p(x 2)2 + y2 = (x 1)2 + (y 1)2

1+i

Re(z)

PAG

Squaring both sides of the equation and

expanding:

ES

|z 2| = |z (1 + i)|

x2 4x + 4 + y2 = x2 2x + 1 + y2 2y + 14x + 4 = 2x 2y + 2y= x1

Method 2 (geometric)

The set S consists of all points in the complex plane that are equidistant from 2 and 1 + i.

AL

In the Cartesian plane, this set corresponds to the perpendicular bisector of the linesegment joining (2, 0) and (1, 1). The midpoint of the line segment is ( 23 , 21 ), and thegradient of the line segment is 1.Therefore the equation of the perpendicular bisector isy

12

= 1(x 23 )

which simplifies to y = x 1.

FIN

Example 31

Sketch the subset of the complex plane defined by each of the following conditions:

The imaginary number i has the property i 2 = 1.

The set of complex numbers is C = { a + bi : a, b R }.

the imaginary part of z is Im(z) = b.

Complex numbers z1 and z2 are equal if and only if Re(z1 ) = Re(z2 ) and Im(z1 ) = Im(z2 ). An Argand diagram is a geometric representation of C.

The modulus of z, denoted by |z|, is the distance from the origin to the point representing z

PAG

in an Argand diagram. Thus |a + bi| = a2 + b2 .

The argument of z is an angle measured anticlockwise about the origin from the positivedirection of the real axis to the line joining the origin to z. The principal value of the argument, denoted by Arg z, is the angle in the interval (, ]. The complex number z = a + bi can be expressedIm(z)in polar form asPz = r(cos + i sin )

Multiplication and division in polar form:

Let z1 = r1 cis 1 and z2 = r2 cis 2 . Then

De Moivres theorem (r cis )n = rn cis(n), where n Z

Conjugate root theorem If a polynomial has real coefficients, then the complex roots

occur in conjugate pairs.

Fundamental theorem of algebra Every non-constant polynomial with complex

coefficients has at least one linear factor in the complex number system.A polynomial of degree n can be factorised over C into a product of n linear factors.If z1 is a solution of z2 = a, where a C, then the other solution is z2 = z1 .The solutions of zn = a, where a C, lie on the circle centred at the origin with12radius |a| n . The solutions are equally spaced around the circle at intervals of.nThe distance between z1 and z2 in the complex plane is |z1 z2 |.

For example, the set z : |z (1 + i)| = 2 is a circle with centre 1 + i and radius 2.

Chapter 4 review 199

Express each of the following in the form a + bi, where a, b R:

Solve each of the following equations for z:

z 2ia (z 2)2 + 9 = 0b=2z + (3 2i)d z4 + 81 = 0

e z3 27 = 0

ES

c z2 + 6z + 12 = 0f 8z3 + 27 = 0

a Show that 2 i is a solution of the equation z3 2z2 3z + 10 = 0. Hence solve

the equation for z.b Show that 3 2i is a solution of the equation x3 5x2 + 7x + 13 = 0. Hence solvethe equation for x C.c Show that 1 + i is a solution of the equation z3 4z2 + 6z 4 = 0. Hence find theother solutions of this equation.

Express each of the following polynomials as a product of linear factors:

a 2x2 + 3x + 2

PAG

b x3 x2 + x 1

c x3 + 2x2 4x 8

If (a + bi)2 = 3 4i, find the possible values of a and b, where a, b R.

Pair each of the transformations given on the left with the appropriate operation on thecomplex numbers given on the right:

AL

a reflection in the real axis

i multiply by 1

b rotation anticlockwise by 90 about O

ii multiply by i

c rotation through 180 about O

iii multiply by i

FIN

d rotation anticlockwise about O through 270

iv take the conjugate

If (a + bi)2 = 24 10i, find the possible values of a and b, where a, b R.

Let w = 1 + i and z = 1 3i.

a Write down:i |w|

13

ii |z|

iii Arg wiv Arg z

wb Hence write down and Arg(wz).z

Express 3 + i in polar form. Hence find 3 + i 7 and express in Cartesian form.Consider the equation z4 2z3 + 11z2 18z + 18 = 0. Find all real values of r for whichz = ri is a solution of the equation. Hence find all the solutions of the equation.

If z = 1 + i is one solution of an equation of the form z4 = a, where a C, then the other

solutions areA 1, 1, 0D 1 + i, 1 i, 1

10

B 1, 1, 1 iE 1 + i, 1 i, 1

C 1 + i, 1 i, 1 i

The square roots of 2 2 3i in polar form are

2

2 A 2 cis , 2 cisB 2 cis , 2 cis3333 2

D 4 cis , 4 cisE 4 cis , 4 cis3333

2

C 4 cis , 4 cis33

PAG

C 3 cis 4

ES

Review

202 Chapter 4: Complex numbers

The zeroes of the polynomial 2x2 + 6x + 7 are and . The value of | | is

105A 5B 2 5C 4 5DE210

Extended-response questions

5

Let z = 4 cisand w = 2 cis.64a Find |z7 | and Arg(z7 ).

AL

FIN

b Show z7 on an Argand diagram.

zc Express in the form r cis .wzd Express z and w in Cartesian form, and hence express in Cartesian form.w 7

e Use the results of d to find an exact value for tan

in the form a + b, where a12and b are rational. 7 f Use the result of e to find the exact value of tan.6

Let v = 2 + i and P(z) = z3 7z2 + 17z 15.

a Show by substitution that P(2 + i) = 0.b Find the other two solutions of the equation P(z) = 0.c Let i be the unit vector in the positive Re(z)-direction and let j be the unit vector inthe positive Im(z)-direction.Let A be the point on the Argand diagram corresponding to v = 2 + i.Let B be the point on the Argand diagram corresponding to 1 2i.

solutions in Cartesian form.b i Plot the two solutions from a on an Argand diagram.ii Find the equation of the circle, with centre the origin, which passes through thesetwo points.iii Find the value of a Z such that the circle passes through (0, a).

in the form (z w)3 = 4 3 4i, where w is a complex number.

a Find the vector XY and the vector XZ.

b Let z1 and z2 be the complex numbers corresponding to the vectors XY and XZ.Find z3 such that z2 = z3 z1 .c By writing z3 in modulusargument form, show that XYZ is half an equilateraltriangle XWZ and give the complex number to which W corresponds.

d The triangle XYZ is rotated through an angle of anticlockwise about Y. Find the3new position of X.3

A regular hexagon LMNPQR has its centre at the origin O and its vertex L at the pointz = 4.

10

11

ES

a Indicate in a diagram the region in the hexagon in which the inequalities |z| 2 and

arg z are satisfied.

33b Find, in the form |z c| = a, the equation of the circle through O, M and R.c Find the complex numbers corresponding to the points N and Q.d The hexagon is rotated clockwise about the origin by 45 . Express in the form r cis the complex numbers corresponding to the new positions of N and Q.1a A complex number z = a + bi is such that |z| = 1. Show that = z.z